# How to simplify arc length problems: expressions under the root

I am having trouble during my studies of arc length by integration in Calculus.

Arc Length formula: $\int\sqrt{1+('y)^2}dx$

My problem values:

$y=\frac{1}{10}x^5+\frac{1}{6}x^-3$

$y'=\frac{5}{10}x^5+(-\frac{3}{6}x^{-4})$ = $\frac{1}{2}x^4-\frac{1}{2}x^-4$ = $\frac{1}{2}(x^4-x^{-4})$

$1+('y)^2=1+(\frac{1}{2}(x^4-x^{-4})^2$

My question is: How does $1+(\frac{1}{2}(x^4-x^{-4})^2$ become $(\frac{1}{2}(x^4+x^{-4})^2$ as seen in the solution?

Is this an obvious algebra step that I am missing? I understand that the soln completes the square, does addition, and achieves sign change, but I don't understand the motive or method. The steps in question can be seen in the linked solution at the bottom of this post. Thank you.

You have $$y'=\frac 12x^4-\frac 12x^{-4}$$ $$\implies y'^2=\frac 14x^8-\frac 12+\frac 14x^{-8}$$ $$\implies 1+y'^2=\frac 14x^8+\frac 12+\frac 14x^{-8}=(\frac 12x^4+\frac12x^{-4})^2$$
\begin {align}1+\frac12\left(x^4-x^{-4}\right)^2&=1+\frac12\left(x^8-2\cdot x^4 \cdot x^{-4}+x^{-8}\right)\\&=1+\frac12(x^8-2+x^{-8})\\ &=-1+\frac12(x^8+2+x^{-8})\\ &=-1+\frac12\left(x^4+x^{-4}\right)^2\end {align} Your expression is missing the $-1$ out front.
• I expanded the square in the first line. When the two terms in the expression to be squared are inverses the middle term is always $2$. In the last line I recompressed the square. Apr 5, 2018 at 14:07
• Between line 2 and line 3 I went from $1+\frac 12(-2)$ to $-1+\frac12(+2)$. Note the interior sign change. Both of them are $0$ so we could just write the expression as $\frac12(x^8+x^{-8})$ if we wanted Apr 5, 2018 at 15:41